The maximum height attained by a projectile when thrown at an angle θ with the horizontal is found to be half the horizontal range. Then θ =

The correct answer is option 4) i.e. tan-1 2

CONCEPT:

• Projectile motion: A projectile is any object that once thrown continues in motion by its own inertia.
  • A projectile motion takes place under the influence of the force of gravity only.

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• The maximum height (h_max) of a projectile is the vertical distance at which the projectile reaches zero vertical velocity.

It is given by the equation:​  h_max = ​\( \frac{u^2\:\sin ^2θ }{2g}\)

• The range (R) of a projectile is the horizontal distance covered by the projectile during its motion. It is given by the equation:

R = \( \frac{u^2\:\sin 2θ }{g}\)

​Where u is the velocity with which the projectile is thrown, θ is the angle at which the projectile is thrown, and g is the acceleration due to gravity

CALCULATION:

Given that:

Maximum height attained = Half the horizontal range

h_max  = \(\frac{1}{2}\)R                              

\( \frac{u^2\:\sin ^2θ }{2g}\) = \(\frac{1}{2}\)\( \frac{u^2\:\sin 2θ }{g}\)

\(sin ^2θ\) = \(sin 2θ\)

\(sin ^2θ\) = \(2sin θ cosθ\)

\(tanθ\) = 2

\(θ=tan^{-1}(2)\)